Subtract:
Question: Subtract: (i) 5xyfrom 12xy (ii) 2a2from 7a2 (iii) 2a bfrom 3a 5b (iv) 2x3 4x2+ 3x+ 5 from 4x3+x2+x+ 6 (v) $\frac{2}{3} y^{3}-\frac{2}{7} y^{2}-5$ from $\frac{1}{3} y^{3}+\frac{5}{7} y^{2}+y-2$ (vi) $\frac{3}{2} x-\frac{5}{4} y-\frac{7}{2} z$ from $\frac{2}{3} x+\frac{3}{2} y-\frac{4}{3} z$ (vii) $x^{2} y-\frac{4}{5} x y^{2}+\frac{4}{3} x y$ from $\frac{2}{3} x^{2} y+\frac{3}{2} x y^{2}-\frac{1}{3} x y$ (viii) $\frac{a b}{7}-\frac{35}{3} b c+\frac{6}{5} a c$ from $\frac{3}{5} b c-\frac{...
Read More →‘If two angles and a side of one triangle
Question: If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent. Is the statement true? Why? Solution: No, because sides must be corresponding sides....
Read More →‘If two sides and an angle of one triangle
Question: If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent. Is the statement true? Why? Solution: No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule....
Read More →The area of a parallelogram is 392 m2.
Question: The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude. Solution: Area of the parallelogram = 392 m2Let the base of the parallelogram bebm.Given:Height of the parallelogram is twice the base.Height = 2bm Area of a parallelogram $=$ Base $x$ Height $\Rightarrow 392=b \times 2 b$ $\Rightarrow 392=2 b^{2}$ $\Rightarrow \frac{392}{2}=b^{2}$ $\Rightarrow 196=b^{2}$ $\Rightarrow b=14$ Base = 14 m Altitude $=2 \times$ Base $...
Read More →The area of a parallelogram is 392 m2.
Question: The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude. Solution: Area of the parallelogram = 392 m2Let the base of the parallelogram bebm.Given:Height of the parallelogram is twice the base.Height = 2bm Area of a parallelogram $=$ Base $x$ Height $\Rightarrow 392=b \times 2 b$ $\Rightarrow 392=2 b^{2}$ $\Rightarrow \frac{392}{2}=b^{2}$ $\Rightarrow 196=b^{2}$ $\Rightarrow b=14$ Base = 14 m Altitude $=2 \times$ Base $...
Read More →The value of the determinant
Question: The value of the determinant $\left|\begin{array}{ccc}a^{2} a 1 \\ \cos n x \cos (n+1) x \cos (n+2) x \\ \sin n x \sin (n+1) x \sin (n+2) x\end{array}\right|$ is independent of (a) $n$ (b) $\mathrm{a}$ (c) $x$ (d) none of these Solution: (a) $n$ Let $A=\mathrm{nx}, B=(\mathrm{n}+1) \mathrm{x}, C=(\mathrm{n}+2) \mathrm{x}$ $\Rightarrow C-B=\mathrm{x}, B-A=\mathrm{x}, C-A=2 \mathrm{x}$ Thus, the given determinant is \begin{tabular}{|lll} $\mathrm{a}^{2}$ $\mathrm{a}$ 1 $\begin{array}{lll...
Read More →In ΔABC and ΔPQR,
Question: In ΔABC and ΔPQR, A = Q and B = R. Which side of ΔPQR should be equal to side BC of ΔABC, so that the two triangles are congruent? Give reason for your answer. Solution: We have given, in ΔABC and ΔPQR, A = Q and B = R Since, two pairs of angles are equal in two triangles. We know that, two triangles will be congruent by AAS rule, if two angles and the side of one triangle are equal to the two angles and the side of other triangle. BC = RP...
Read More →The area of a parallelogram is 392 m2.
Question: The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude. Solution: Area of the parallelogram = 392 m2Let the base of the parallelogram bebm.Given:Height of the parallelogram is twice the base.Height = 2bm Area of a parallelogram $=$ Base $x$ Height $\Rightarrow 392=b \times 2 b$ $\Rightarrow 392=2 b^{2}$ $\Rightarrow \frac{392}{2}=b^{2}$ $\Rightarrow 196=b^{2}$ $\Rightarrow b=14$ Base = 14 m Altitude $=2 \times$ Base $...
Read More →In ΔABC and ΔPQR,
Question: In ΔABC and ΔPQR, A = Q and B =R. Which side of ΔPQR should be equal to side AB of ΔABC, so that the two triangles are congruent? Give reason for your answer. Solution: We have given, in $\triangle A B C$ and $\triangle P Q R, \angle A=\angle Q$ and $\angle B=\angle R$ Since, AB and QR are included between equal angles. Hence, the side of ΔPQR is QR which should be equal to side AB of ΔABC, so that the triangles are congruent by the rule ASA....
Read More →Add the following algebraic expressions:
Question: Add the following algebraic expressions: (i) 3a2b, 4a2b, 9a2b (ii) $\frac{2}{3} a, \frac{3}{5} a,-\frac{6}{5} a$ (iii) 4xy2 7x2y, 12x2y 6xy2, 3x2y+5xy2 (iv) $\frac{3}{2} a-\frac{5}{4} b+\frac{2}{5} c, \frac{2}{3} a-\frac{7}{2} b+\frac{7}{2} c, \frac{5}{3} a+\frac{5}{2} b-\frac{5}{4} c$ (v) $\frac{11}{2} x y+\frac{12}{5} y+\frac{13}{7} x,-\frac{11}{2} y-\frac{12}{5} x-\frac{13}{7} x y$ (vi) $\frac{7}{2} x^{3}-\frac{1}{2} x^{2}+\frac{5}{3}, \frac{3}{2} x^{3}+\frac{7}{4} x^{2}-x+\frac{1}{...
Read More →Solve this
Question: Let $\left|\begin{array}{ccc}x 2 x \\ x^{2} x 6 \\ x x 6\end{array}\right|=a x^{4}+b x^{3}+c x^{2}+d x+e$ Then, the value of $5 a+4 b+3 c+2 d+e$ is equal to (a) 0 (b) $-16$ (c) 16 (d) none of these Solution: (d) none of these $\Delta=\mid \begin{array}{lll}x 2 x\end{array}$ $x^{2} \quad x \quad 6$ $\begin{array}{lll}x x 6\end{array}$ $=x \mid x \quad 6$ $\begin{array}{ll}x 6\left|-x^{2}\right| 2 x \\ x 6|+x| 2 x\end{array}$ $\begin{array}{ll}x 6\end{array}$ [Expanding along $C_{1}$ ] $...
Read More →In ΔABC and ΔDEF, AB = FD and ∠A = ∠D.
Question: In ΔABC and ΔDEF, AB = FD and A = D. The two triangles will be congruent by SAS axiom, if (a) BC = EF (b) AC = DE (c)AC=EF (d) BC = DE Solution: (b) Given, in ΔABC and ΔDEF, AB = DF and A = D We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle. AC = DE...
Read More →The adjacent sides of a parallelogram are 32 cm and 24 cm.
Question: The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find hte distance between the shorter sides. Solution: Longer side = 32 cmShorter side = 24 cmLet the distance between the shorter sides bexcm. Area of a parallelogram = Longer side $\times$ Distance between the longer sides $=$ Shorter side $\times$ Distance between the shorter sides or, $32 \times 17.4=24 \times x$ or, $x=\frac{32 \times 17.4}{24}=23.2 \mathrm{~cm}$ Distanc...
Read More →Find the area of a parallelogram with base equal to 25 cm
Question: Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm. Solution: Given:Base = 25 cmHeight = 16.8 cm $\therefore$ Area of the parallelogram $=$ Base $\times$ Height $=25 \mathrm{~cm} \times 16.8 \mathrm{~cm}=420 \mathrm{~cm}^{2}$...
Read More →Which of the following is not correct in a given determinant of A,
Question: Which of the following is not correct in a given determinant of $A$, where $A=\left[a_{i}\right]_{3 \times 3}$. (a) Order of minor is less than order of the det (A) (b) Minor of an element can never be equal to cofactor of the same element (c) Value of determinant is obtained by multiplying elements of a row or column by corresponding cofactors (d) Order of minors and cofactors of elements of $A$ is same Solution: (b) Minor of an element can never be equal to the cofactor of the same e...
Read More →Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm,
Question: Find the area of the quadrilateralABCDin whichAB= 42 cm,BC= 21 cm,CD= 29 cm,DA= 34 cm and diagonalBD= 20 cm. Solution: Area of $\triangle A B D=\sqrt{s(s-a)(s-b)(s-c)}$ $s=\frac{1}{2}(a+b+c)$ $s=\frac{42+20+34}{2}$ $s=48 \mathrm{~cm}$ Area of $\triangle A B D=\sqrt{48(48-42)(48-20)(48-34)}$ $=\sqrt{48 \times 6 \times 28 \times 14}$ $=\sqrt{112896}$ $=336 \mathrm{~cm}^{2}$ Area of $\triangle B D C=\sqrt{s(s-a)(s-b)(s-c)}$ $s=\frac{1}{2}(a+b+c)$ $s=\frac{21+20+29}{2}$ $s=35 \mathrm{~cm}$...
Read More →In ΔABC and ΔPQR, if AB = AC,
Question: In ΔABC and ΔPQR, if AB = AC, C =P and B = Q, then the two triangles are (a) isosceles but not congruent (b) isosceles and congruent (c) congruent but not isosceles (d) Neither congruent nor isosceles Solution: (a) $\ln \triangle A B C$, $A B=A C$ [given] $\Rightarrow \quad \angle C=\angle B \quad$ [angles opposite to equal sides are equal] So, $\triangle A B C$ is an isosceles triangle. But it is given that, $\angle B=\angle Q$ $\angle C=\angle P$ $\therefore \quad \angle P=\angle Q \...
Read More →Solve this
Question: If $A=\left|\begin{array}{lll}a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{31} a_{32} a_{33}\end{array}\right|$ and $C_{i j}$ is cofactor of $a_{i j}$ in $A$, then value of $|A|$ is given by (a) $a_{11} C_{31}+a_{12} C_{32}+a_{13} C_{33}$ (b) $a_{11} C_{11}+a_{12} C_{21}+a_{13} C_{31}$ (c) $a_{21} C_{11}+a_{22} C_{12}+a_{23} C_{13}$ (d) $a_{11} C_{11}+a_{21} C_{21}+a_{13} C_{31}$ Solution: (d) $a_{11} C_{11}+a_{21} C_{21}+a_{13} C_{31}$ Properties of determinants state that if $A...
Read More →In ΔPQR, if ∠R > ∠Q, then
Question: In ΔPQR, if R Q, then (a) QR PR (b) PQ PR (c) PQ PR (d) QR PR Solution: (b) Given, $\angle R\angle Q$ $\Rightarrow \quad P QP R \quad$ [side opposite to greater angle is longer]...
Read More →Evaluate the following:
Question: Evaluate the following: (i) 102 106 (ii) 109 107 (iii) 35 37 (iv) 53 55 (v) 103 96 (vi) 34 36 (vii) 994 1006 Solution: (i) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$ $102 \times 106$ $=(100+2)(100+6)$ $=100^{2}+(2+6) 100+2 \times 6$ $=10000+800+12$ $=10812$ (ii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$ $109 \times 107$ $=(100+9)(100+7)$ $=100^{2}+(9+7) 100+9 \times 7$ $=10000+1600+63$ $=11663$ (iii) Here, we will use the identity $(x+a)(x+b)=x^{2...
Read More →Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm,
Question: Find the perimeter and area of the quadrilateralABCDin whichAB= 17 cm,AD= 9 cm,CD= 12 cm, ACB= 90 andAC= 15 cm. Solution: In the right-angledΔACB: $A B^{2}=B C^{2}+A C^{2}$ $\Rightarrow 17^{2}=B C^{2}+15^{2}$ $\Rightarrow 17^{2}-15^{2}=B C^{2}$ $\Rightarrow 64=B C^{2}$ $\Rightarrow B C=8 \mathrm{~cm}$ Perimeter $=A B+B C+C D+A D$ $=17+8+12+9$ = 46 cm Area of $\Delta A B C=\frac{1}{2}(b \times h)$ $=\frac{1}{2}(8 \times 15)$ $=60 \mathrm{~cm}^{2}$ In $\Delta A D C:$ $A C^{2}=A D^{2}+C D...
Read More →If two sides of a triangle are of lengths 5 cm
Question: If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be (a) 3.6 cm (b) 4.1 cm (c) 3.8 cm (d) 3.4 cm Thinking Process Use the condition that, sum of any two sides of a triangle is greater than third side and difference of any two sides is less than the third side. Solution: (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed ...
Read More →Which of the following is not correct?
Question: Which of the following is not correct? (a) $|A|=\left|A^{T}\right|$, where $A=\left[a_{i j}\right]_{3 \times 3}$ (b) $|k A|=\left|k^{3}\right|$, where $A=\left[a_{i j}\right]_{3 \times 3}$ (c) If $A$ is a skew-symmetric matrix of odd order, then $|A|=0$ (d) $\left|\begin{array}{ll}a+b c+d \\ e+f g+h\end{array}\right|=\left|\begin{array}{ll}a c \\ e g\end{array}\right|+\left|\begin{array}{ll}b d \\ f h\end{array}\right|$ Solution: (d) $\left|\begin{array}{ll}a+b c+d \\ e+f g+h\end{array...
Read More →It is given that ΔABC = ΔFDE and AB = 5 cm,
Question: It is given that ΔABC = ΔFDE and AB = 5 cm, B = 40 and A = 80, then Which of the following is true? (a) DF = 5 cm, F = 60 (b) DF = 5 cm, E =60 (c) DE = 5 cm, E = 60 (d) DE = 5 cm, D = 40 Solution: (b) Given, $\triangle A B C \cong \triangle F D E$ and $A B=5 \mathrm{~cm}, \angle B=40^{\circ}, \angle A=80^{\circ}$ Since, $\triangle F D E \cong \triangle A B C$ $\therefore \quad D F=A B$ [by CPCT] $D F=5 \mathrm{~cm}$ and $\angle E=\angle C$ [by CPCT] $\Rightarrow \quad \angle E=\angle C...
Read More →If D is a point on the side BC of a ΔABC
Question: If D is a point on the side BC of a ΔABC such that AD bisects BAC. Then, (a) BD = CD (b) BA BD (c) BD BA (d)CD CA Thinking Process (i) Firstly, use the property, exterior angle of a triangle is greater than interior opposite angle. (ii) Secondly, use the property that in a triangle, the side opposite to the greater angle is longer. Solution: (b) Given, $\triangle A B C$ such that $A D$ bisects $\angle B A C$ $\therefore$ $\angle B A D=\angle C A D$ $\ldots(i)$ In $\triangle A C D, \ang...
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